Question

Please explain the process and equation to get the answer

Answer 1

The shell's horizontal position
`x` and vertical position
`y` are given by

`x=v_0\cos55.0^\circ\,t`

`y=v_0\sin55.0^\circ\,t-\frac g2t^2`

where
`v_0` is the given speed of 1.70 x 10^4 m/s, and
`g` is the acceleration due to gravity (taken here to be 9.80 m/s^2).

To find the horizontal range, you can use the range formula,

`x_(\rm max)=\frac{{v_0}^2\sin2\theta}g`

with
`\theta` being the angle at which the shell is fired.

Alternatively, we can work backwards and deal with part (b) first:

(b) The time spent in the air is the time it takes for the shell to reach the ground. To find that, you solve for
`t` in
`y=0`:

`v_0\sin55.0^\circ\,t-\frac g2t^2=0\implies t\approx284\,\rm s`

(a) After this time, the shell will have traveled horizontally

`x=v_0\cos55.0^\circ(284\,\rm s)\approx2.77*10^5\,\rm m`